Calculate Moles of oxygen needed to react with 4.21 mol of Al. What mass of Al2O3 should be produced?
4 Al (s) + 3 O2 (g) → 2 Al2O3 (s)
Calculate Moles of oxygen needed to react with 4.21 mol of Al. What mass of Al2O3...
1. When 15.4 g of aluminum (26.98 g/mol) reacts with 14.8 g of oxygen gas (32.00 g/mol), what is the maximum mass (in grams) of aluminum oxide (101.96 g/mol) that can be produced? 4 Al(s) + 3 O2(g) → 2 Al2O3(s) 2. If 14.1 moles of Cu and 48.7 moles of HNO3 are allowed to react, how many moles of excess reactant will remain if the reaction goes to completion? Cu + 8 HNO3 → 3Cu(NO3)2 + 2NO + 4H2O...
How many grams of Al2O3 are produced when 38.9 grams of O2 react completely with Al? 4 Al (s) + 3 O2 (g) → 2 Al2O3 (s)
How many moles of oxygen (O2) are needed to react with 12.50 moles of C2H2 in the reaction? 2C2H2 + 5O2 ⟶4 CO2 + 2 H2O C2H2 = 26.038 g/mol; O2 = 31.998 g/mol; CO2 = 44.01 g/mol; H2O = 18.01 g/mol Group of answer choices 42.9 g 31.3 23.7 5.00
What is the amount (in moles) of excess reactant remaining if all of the limiting reactant completely reacts when 24.9 mol of aluminum and 27.9 moles of oxygen gas react? 4 Al (s) + 3 O2(g) → 2 Al2O3(s)
Aluminum oxide (Al2O3) is produced according to the following equation. 4 Al(s) + 3 O2(g) → 2 Al2O3(s) If the reaction occurs with an 82.4% yield, what mass of aluminum should be reacted with excess oxygen to produce 45.0 grams of Al2O3? a. 54.6 g Al b. 37.9 g Al c. 35.1 g Al d. 23.8 g Al e. 28.9 g Al How do we solve this problem
6. If 5.00 mol of nitrogen gas and 5.00 mol of oxygen gas react, what is the limiting reactant and how many moles of NO are produced from the reaction? N2(g) + O2 (g) -> 2 NO(g)
5. If 1.00 mol of nitrogen monoxide gas and 1.00 mol of oxygen gas react, what is the limiting reactant and how many moles of NO are produced from the reaction? 2 NO(g) + O2(g) -> 2 NO(g)
A. How many moles of Al can be produced from 10.87 g of Ag? Al(NO3)3(aq) + 3 Ag(s) → Al(s) + 3 AgNO3(aq) B. What mass of O2 can be generated by the decomposition of 100.0 g of NaClO3? 2NaClO3(s) → 2 NaCl(s) + 3 O2(g) C. What mass of Fe is generated when 100.0 g of Al are reacted? Fe2O3(s) + 2 Al(s) → 2 Fe(s) + Al2O3(s)
Part A How many moles of O2 are needed to react with 1.95 mol of C2H2? Part BHow many moles of CO2 are produced when 3.3 mol of C2H2 react?Part CHow many moles of C2H2 are required to produce 0.61 mol of H2O? Part D How many moles of CO2 are produced when 0.145 mol of O2 reacts?
help me do the correct math 25. (12 pts) Aluminum oxide forms when aluminum-reacts with oxygen. Balance the following chemical equation (the molar mass of wluminum oxide is 101.96 g/mol): 4 Al+ 3 02 2 Al2O3 62.79 g of aluminum metal is allowed to react. How many moles of aluminum is this? 52.79 gray.101.96 glmolo = 6402 moles Al How many moles of Al2O3 can be produced from the previously calculated moles of aluminum? 23 6402 x 6o2to² 3 =31858...